Problem 1 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Consider the following situation. Suppose we have two fair dice, \(D_{1}\) with 5 red sides and 1 white side and \(D_{2}\) with 1 red side and 5 white sides. We pick one of the dice randomly, and throw it repeatedly until red comes up for the first time. With the same die this experiment is repeated two more times. Suppose the following happens: First experiment: first red appears in 3rd throw Second experiment: first red appears in 5 th throw Third experiment: first red appears in 4 th throw. Show that for die \(D_{1}\) this happens with probability \(5.7424 \cdot 10^{-8}\), and for die \(D_{2}\) the probability with which this happens is \(8.9725 \cdot 10^{-4}\). Given these probabilities, which die do you think we picked?

Step-by-Step Solution

Answer

Die \(D_2\) was likely chosen, as it has a higher probability of all results occurring.

Step 1: Calculate Probability for Die D_1

The probability of getting a red face on die \(D_1\) is \(\frac{5}{6}\), and the probability of getting a white face is \(\frac{1}{6}\). For each experiment: **First experiment:** Red appears on the 3rd throw. The first two throws must be white and the third red, which is \(\left(\frac{1}{6}\right)^2 \cdot \frac{5}{6}\). **Second experiment:** Red appears on the 5th throw, which is \(\left(\frac{1}{6}\right)^4 \cdot \frac{5}{6}\). **Third experiment:** Red appears on the 4th throw, which is \(\left(\frac{1}{6}\right)^3 \cdot \frac{5}{6}\). The total probability for all three experiments is the product of these probabilities:\[P(D_1) = \left(\frac{1}{6}\right)^2 \cdot \frac{5}{6} \cdot \left(\frac{1}{6}\right)^4 \cdot \frac{5}{6} \cdot \left(\frac{1}{6}\right)^3 \cdot \frac{5}{6}\]

Step 2: Calculate Numerical Result for Die D_1

Using a calculator, simplify the expression from Step 1:\[P(D_1) = \left(\frac{1}{6}\right)^9 \cdot \left(\frac{5}{6}\right)^3 = \frac{1}{60466176} \cdot \frac{125}{216} \approx 5.7424 \cdot 10^{-8}\]

Step 3: Calculate Probability for Die D_2

The probability of getting a red face on die \(D_2\) is \(\frac{1}{6}\), and the probability of getting a white face is \(\frac{5}{6}\). For each experiment:**First experiment:** Red appears on the 3rd throw. The first two throws must be white, and the third red, which is \(\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}\). **Second experiment:** Red appears on the 5th throw, which is \(\left(\frac{5}{6}\right)^4 \cdot \frac{1}{6}\). **Third experiment:** Red appears on the 4th throw, which is \(\left(\frac{5}{6}\right)^3 \cdot \frac{1}{6}\). The total probability for all three experiments is the product of these probabilities:\[P(D_2) = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^4 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^3 \cdot \frac{1}{6}\]

Step 4: Calculate Numerical Result for Die D_2

Using a calculator, simplify the expression from Step 3:\[P(D_2) = \left(\frac{5}{6}\right)^9 \cdot \left(\frac{1}{6}\right)^3 = \frac{1953125}{10077696} \approx 8.9725 \cdot 10^{-4}\]

Step 5: Compare Probabilities and Make a Decision

Since the probability \(8.9725 \cdot 10^{-4}\) for die \(D_2\) is significantly higher than the probability \(5.7424 \cdot 10^{-8}\) for die \(D_1\), it is much more likely that die \(D_2\) was chosen for the experiment.

Key Concepts

Understanding Random Variables

In probability theory, random variables are fundamental. They are used to model outcomes of random experiments. A random variable is essentially a function that assigns a number to each outcome of a random process.
In our dice problem, we have two dice, each acting as a random variable. Die \( D_{1} \) and die \( D_{2} \) have different probabilities for landing on a red or white side:

• For \( D_{1} \), there are 5 red sides and 1 white side, meaning the probability \( P(Red) = \frac{5}{6} \) and \( P(White) = \frac{1}{6} \).
• For \( D_{2} \), it is opposite; \( P(Red) = \frac{1}{6} \) and \( P(White) = \frac{5}{6} \).

This shows how a random variable can represent different distributions depending on its conditions, which is key in calculating probabilities in various scenarios.

Diving Into Conditional Probability

Conditional probability is the probability of an event occurring, given that another event has occurred. It is fundamental in our dice scenario, where initially choosing one die affects the outcomes of subsequent trials.
If we select die \( D_{1} \), the outcomes like 'first red in the 3rd throw' depend on the condition of having \( D_{1} \). The probability for this can be calculated as \( \left(\frac{1}{6}\right)^2 \cdot \frac{5}{6} \). Similarly, for die \( D_{2} \), it is \( \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6} \). Each of these calculations assumes a condition initially set by the die.
Conditional probabilities in repetitive experiments help us in determining which die might have been used, by comparing calculated outcomes under each condition.

Connecting with Mathematical Statistics

Mathematical statistics uses probability theory to make inferences about populations based on sample data. In our dice problem, after calculating probabilities for both dies \( D_{1} \) and \( D_{2} \), we compare these to infer which die was likely used. This is the decision-making part of statistics.
The probabilities calculated: \( 5.7424 \cdot 10^{-8} \) for \( D_{1} \) and \( 8.9725 \cdot 10^{-4} \) for \( D_{2} \), are interpreted to conclude that \( D_{2} \) is more likely since its probability is significantly higher.

• This comparison reflects the essence of hypothesis testing, where hypotheses about which die was used are tested against the experimental outcome probabilities.

Thus, mathematical statistics allows us to use calculated probabilities to derive meaningful conclusions from random experiments.